Insulin is a polypeptide hormone consisting of two chains A and B linked by two inter‑chain disulfide bonds and one intra‑chain bond in the A chain. Explain how reduction of these disulfide bonds affects insulin’s ability to bind to its receptor. Include a brief description of the structural consequence and the functional outcome.
Answer
Reduction breaks disulfide bridges, unfolding insulin and abolishing its receptor binding ability.
The correct answer is Reduction breaks disulfide bridges, unfolding insulin and abolishing its receptor binding ability. — Conclude that the unfolded insulin cannot adopt the conformation required for high‑affinity receptor binding, thus losing its biological activity.
Step-by-step solution
- Describe insulin’s native structure: two chains linked by disulfide bonds that stabilize its folded conformation.
- Explain that reducing agents (e.g., dithiothreitol) cleave disulfide bridges, converting –S–S– bonds to free –SH groups.
- State that breaking these bonds disrupts the tertiary structure, leading to an unfolded or partially folded protein.
- Conclude that the unfolded insulin cannot adopt the conformation required for high‑affinity receptor binding, thus losing its biological activity.
Common mistake: Believing reduction has no effect on insulin’s function
Concept tested: Disulfide bridges stabilize protein folding
Published · How this solution was prepared and checked
Related questions
- 1 During DNA replication, a DNA polymerase synthesises a complementary strand. If the template strand contains 30 nucleotides, how many phosphodiester bonds are formed in the new strand?
- 2 A 200 g sample of soybean oil contains 20 % triglycerides. When it is heated at 200 °C for 2 h, 15 % of the triglycerides hydrolyze to free fatty acids. How many grams of free fatty acids are formed?
- 3 At 0 °C, glycine (NH₂CH₂COOH) is treated with excess nitrous acid (HNO₂). What gas is evolved?
- 4 In an electrolytic cell, explain why metal ions are reduced at the cathode even though the cathode is the negative electrode.
- 5 Explain why a 1 mol kg^-1 aqueous solution of sucrose shows a lower vapour pressure than that predicted by Raoult's law. In your answer discuss the role of the non‑volatile solute, solute–solvent interactions, and the activity coefficient.
- 6 A transition metal ion is d^6 in a high‑spin octahedral complex. The octahedral crystal‑field splitting energy Δ_o is 18,000 cm^-1 and the electron‑pairing energy P is 12,000 cm^-1. Calculate the crystal‑field stabilization energy (CFSE) for the high‑spin and low‑spin possibilities and decide which spin state is favoured.